Find the sum of all the terms in the nth row of the series given below.
1 2
3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18 19 20
..........................
............................
(so on)
Examples:
Input : n = 2 Output : 18 terms in 2nd row and their sum sum = (3 + 4 + 5 + 6) = 18 Input : n = 4 Output : 132
Naive Approach: Using two loops. Outer loop executes for i = 1 to n times. Inner loop executes for j = 1 to 2 * i times. Counter variable k to keep track of the current term in the series. When i = n, the values of k are accumulated to the sum.
Time Complexity: O(k), where k is the total number of terms from the beginning till the end of the nth row.
Efficient Approach: The sum of all the terms in the nth row can be obtained by the formula:
Sum(n) = n * (2 * n2 + 1)
The proof for the formula is given below:
Prerequisite:
- Sum of n terms of an Arithmetic Progression series with a as the first term and d as the common difference is given as:
Sum = (n * [2*a + (n-1)*d]) / 2
- Sum of 1st n natural numbers is given as:
Sum = (n * (n + 1)) / 2
Proof:
Let the number of terms from the beginning
till the end of the nth row be p.
Here p = 2 + 4 + 6 + .....n terms
For the given AP series, a = 2, d = 2.
Using the above formula for the sum of
n terms of the AP series, we get,
p = n * (n + 1)
Similarly, let the number of terms from the
beginning till the end of the (n-1)th row be q.
Here q = 2 + 4 + 6 + .....n-1 terms
For the given AP series, a = 2, d = 2.
Using the above formula for the sum of
n-1 terms of the AP series, we get,
q = n * (n - 1)
Now,
Sum of all the terms in the nth row
= sum of 1st p natural numbers -
sum of 1st q natural numbers
= (p * (p + 1)) / 2 - (q * (q + 1)) / 2
Substituting the values of p and q and then solving
the equation, we will get,
Sum of all the terms in the nth row = n * (2 * n2 + 1)
C++
// C++ implementation to find the sum of all the // terms in the nth row of the given series #include <bits/stdc++.h> using namespace std; // function to find the required sum int sumOfTermsInNthRow(int n) { // sum = n * (2 * n^2 + 1) int sum = n * (2 * pow(n, 2) + 1); return sum; } // Driver program to test above int main() { int n = 4; cout << "Sum of all the terms in nth row = " << sumOfTermsInNthRow(n); return 0; } |
Java
// Java implementation to find the sum of all the // terms in the nth row of the given series import static java.lang.Math.pow; class Test { // method to find the required sum static int sumOfTermsInNthRow(int n) { // sum = n * (2 * n^2 + 1) int sum = (int)(n * (2 * pow(n, 2) + 1)); return sum; } // Driver method public static void main(String args[]) { int n = 4; System.out.println("Sum of all the terms in nth row = " + sumOfTermsInNthRow(n)); } } |
Python3
# Python 3 implementation to find # the sum of all the terms in the # nth row of the given series from math import pow # function to find the required sum def sumOfTermsInNthRow(n): # sum = n * (2 * n^2 + 1) sum = n * (2 * pow(n, 2) + 1) return sum # Driver Code if __name__ == '__main__': n = 4 print("Sum of all the terms in nth row =", int(sumOfTermsInNthRow(n))) # This code is contributed # by Surendra_Gangwar |
C#
// C# implementation to find the sum of all the // terms in the nth row of the given series using System; class Test { // method to find the required sum static int sumOfTermsInNthRow(int n) { // sum = n * (2 * n^2 + 1) int sum = (int)(n * (2 * Math.Pow(n, 2) + 1)); return sum; } // Driver method public static void Main() { int n = 4; Console.Write("Sum of all the terms in nth row = " + sumOfTermsInNthRow(n)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP implementation to find // the sum of all the terms in // the nth row of the given series // function to find the required sum function sumOfTermsInNthRow($n) { // sum = n * (2 * n^2 + 1) $sum = $n * (2 * pow($n, 2) + 1); return $sum; } // Driver Code $n = 4; echo "Sum of all the terms in nth row = ", sumOfTermsInNthRow($n); // This code is contributed by ajit ?> |
Output:
Sum of all the terms in nth row = 132
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